Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $n = \dfrac{r^2 - 9}{-5r^3 - 10r^2 + 15r} \div \dfrac{-r - 1}{r^2 - r} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $n = \dfrac{r^2 - 9}{-5r^3 - 10r^2 + 15r} \times \dfrac{r^2 - r}{-r - 1} $ First factor out any common factors. $n = \dfrac{r^2 - 9}{-5r(r^2 + 2r - 3)} \times \dfrac{r(r - 1)}{-(r + 1)} $ Then factor the quadratic expressions. $n = \dfrac {(r + 3)(r - 3)} {-5r(r + 3)(r - 1)} \times \dfrac {r(r - 1)} {-(r + 1)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac { (r + 3)(r - 3) \times r(r - 1)} { -5r(r + 3)(r - 1) \times -(r + 1)} $ $n = \dfrac {r(r + 3)(r - 3)(r - 1)} {5r(r + 3)(r - 1)(r + 1)} $ Notice that $(r + 3)$ and $(r - 1)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac {r\cancel{(r + 3)}(r - 3)(r - 1)} {5r\cancel{(r + 3)}(r - 1)(r + 1)} $ We are dividing by $r + 3$ , so $r + 3 \neq 0$ Therefore, $r \neq -3$ $n = \dfrac {r\cancel{(r + 3)}(r - 3)\cancel{(r - 1)}} {5r\cancel{(r + 3)}\cancel{(r - 1)}(r + 1)} $ We are dividing by $r - 1$ , so $r - 1 \neq 0$ Therefore, $r \neq 1$ $n = \dfrac {r(r - 3)} {5r(r + 1)} $ $ n = \dfrac{r - 3}{5(r + 1)}; r \neq -3; r \neq 1 $